# Filters and Filter Bases

## Filters

Definition: Let $E$ be a set. A Filter is a collection $\mathscr{F}$ of nonempty subsets of $E$ that satisfy the following properties:(1) If $A, B \in \mathscr{F}$ then $A \cap B \in \mathscr{F}$.(2) If $A \in \mathscr{F}$ and $A \subseteq B$ then $B \in \mathscr{F}$. |

*So a filter is a collection $\mathscr{F}$ of nonempty subsets of $E$ for which intersections of sets in the filter are still contained in the filter, and for which sets containing sets in the filter are also in the filter.*

Let $E$ be a set. Given any subset $A$ of $E$, the collection:

(1)i.e., the collection of all subsets of $E$ which contain $A$ is a filter. Indeed, if $B_1, B_2 \in \mathscr{F}$ then since $A \subseteq B_1$ and $A \subseteq B_2$ we have that $A \subseteq B_1 \cap B_2$ so that $B_1 \cap B_2 \in \mathscr{F}$. Furthermore, if $B_1 \in \mathscr{F}$ and $B_1 \subseteq B_2$ then $A \subseteq B_2$ so that $B_2 \in \mathscr{F}$.

For another example, let $E$ be a topological space. Given a point $a \in E$, the collection:

(2)is a filter. Indeed, if $B_1, B_2 \in \mathscr{F}$ then there exists open sets $U_1, U_2$ with $a \in U_1 \subseteq B_1$ and $a \in U_2 \subseteq B_2$. Then $a \in U_1 \cap U_2 \subseteq B_1 \cap B_2$, so that $B_1 \cap B_2$ is a neighbourhood of $a$ and $B_1 \cap B_2 \in \mathscr{F}$. Furthermore, if $B_1 \in \mathscr{F}$ and $B_1 \subseteq B_2$ then if $U$ is an open set such that $a \in U \subseteq B_1$ then $a \in U \subseteq B_2$ so that $B_2 \in \mathscr{F}$.

Definition: Let $E$ be a set and let $\mathscr{F}$ be a filter. A Refinement of $\mathscr{F}$ is another filter $\mathscr{G}$ such that $\mathscr{F} \subseteq \mathscr{G}$, and we say that $\mathscr{G}$ is Finer than $\mathscr{F}$. |

*In other words, a finer filter is one that contains more sets.*

## Filter Bases

Definition: Let $E$ be a set. A Filter Base is a nonempty collection $\mathscr{B}$ of nonempty subsets of $E$ with the property that:(1) If $A, B \in \mathscr{B}$ then there exists $C \in \mathscr{B}$ such that $C \subseteq A \cap B$.The Filter Generated by $\mathscr{B}$ is the filter consisting of all sets which contain a set of $\mathscr{B}$. |

Note that if $\mathscr{B}$ is a filter base and if $\mathscr{F}$ is the filter generated by $\mathscr{B}$ then $\mathscr{F}$ is indeed a filter. Let $A, B \in \mathscr{F}$. Then there exists $U_1, U_2 \in \mathscr{B}$ such that $U_1 \subseteq A$ and $U_2 \subseteq B$. But since $\mathscr{B}$ is a filter base, there exists a $V \in \mathscr{B}$ such that $V \subseteq U_1 \cap U_2 \subseteq A \cap B$. Thus $A \cap B$ contains a set of $\mathscr{B}$ and so $A \cap B \in \mathscr{F}$.

Now let $A \in \mathscr{F}$ and $A \subseteq B$. Then there exists a $U \in \mathscr{B}$ such that $U \subseteq A$. But then $U \subseteq B$. So $B$ contains a set of $\mathscr{B}$ and so $B \in \mathscr{B}$. So indeed, the filter generated by a filter base is a filter.

Proposition 1: Let $E$ and $F$ be sets and let $f : E \to F$. If $\mathscr{B}$ is a filter base in $E$ then $f(\mathscr{B})$ is a filter base in $F$. |

*Here we define $f(\mathscr{B}) := \{ f(A) : A \in \mathscr{B} \}$.*

**Proof:**Let $\mathscr{B}$ be a filter base in $E$. Let $U, V \in f(\mathscr{B})$. Then there exists $A, B \in \mathscr{B}$ such that $U = f(A)$ and $V = f(B)$.

- Since $\mathscr{B}$ is a filter base in $E$ and $A, B \in \mathscr{B}$ there exists a $C \in \mathscr{B}$ such that $C \subseteq A \cap B$. Then:

- So $f(C) \in f(\mathscr{B})$ is such that $f(C) \subseteq f(A) \cap f(B)$, so $f(\mathscr{B})$ is a filter base in $F$.